package  main.java.leetcode.editor.cn;
//2022-04-12 19:46:57
//给定一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。 
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// 示例 1： 
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//输入：head = [1,2,3,4,5], n = 2
//输出：[1,2,3,5]
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// 示例 2： 
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//输入：head = [1], n = 1
//输出：[]
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// 示例 3： 
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//输入：head = [1,2], n = 1
//输出：[1]
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// 提示： 
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// 链表中结点的数目为 sz 
// 1 <= sz <= 30 
// 0 <= Node.val <= 100 
// 1 <= n <= sz 
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// 进阶：能尝试使用一趟扫描实现吗？ 
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// 注意：本题与主站 19 题相同： https://leetcode-cn.com/problems/remove-nth-node-from-end-of
//-list/ 
// Related Topics 链表 双指针 
// 👍 35 👎 0

class SLwz0R {
    public static void main(String[] args) {
        //创建该题目的对象方便调用
        Solution solution = new SLwz0R().new Solution();
    }
    public class ListNode {
        int val;
        ListNode next;
        ListNode() {}
        ListNode(int val) { this.val = val; }
        ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }
    //leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null){
            return null;
        }
        ListNode dummyHead = new ListNode();
        dummyHead.next = head;
        ListNode fastIndex = dummyHead;
        ListNode slowIndex = dummyHead;
        for(int i = 0;i<n;i++){
            fastIndex = fastIndex.next;
        }
        while (fastIndex.next != null){
            slowIndex = slowIndex.next;
            fastIndex = fastIndex.next;
        }
        slowIndex.next = slowIndex.next.next;

        return dummyHead.next;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}
